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The terms AD and BC are used to denote years in the Gregorian calendar. AD stands for "Anno Domini," which is Latin for "in the year of our Lord," referring to the years after the birth of Jesus Christ. BC stands for "Before Christ," indicating the years before his birth. Together, they create a chronological framework that has been widely used in Western history.
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What is the reason to ab plus bd equals ad?
The equation ( ab + bd = ad ) can be understood through factoring. The left side can be factored as ( b(a + d) ), which shows that it represents the product of ( b ) and the sum ( (a + d) ). However, this equation is not generally true unless specific conditions about the variables ( a ), ( b ), and ( d ) are met. In standard algebra, for ( ab + bd ) to equal ( ad ), we would require ( b = 0 ) or ( a + d = 0 ).
Have a rectangle bdfg with point a on side bd so that ba equals 16 cm and ad equals 8 cm what fraction of the area of the rectangle is inside triangle baf?
ba=(16/(16+8))bd=(16/24)bd=(2/3)bd area of the rectangle = bd*bf area of triangle = (2/3)bd*bf/2=(1/3)*bd*bf 1/3
Why is a over b plus c over d equal ad plus bc divided by bd always valid?
Because it can be derived from a large number of more basic principles applied to numbers. Since a/b and c/d are defined, then b and d are non-zero. Then a/b = ad/bd since multiplying the numerator and denominator of a fraction by a non-zero number leaves it unchanged. Similarly, c/d = bc/bd [These steps implicitly assume the commutative property of multiplication ie there is no difference between premultiplication and postmultiplication] Another property of numbers: If x = p and y = q then x+y = p+q [equals added to equals result in equals]. So a/b + c+d = ad/bd + bc/bd and finally, using the distributive property of multiplication [by 1/bd] over addition, = (ad + bc)/bd Maybe you wish you hadn't asked!
Does distributive property apply to complex numbers?
Yes, for example (a + bi)(c + di) = ac + adi + bic + bidi, and commutative property works as well --> ac + adi + bci + bdi² --> ac + (ad + bc)i + bd(-1) = (ac - bd) + (ad + bc)i
What does a over b plus c over d equal?
The expression ( \frac{a}{b} + \frac{c}{d} ) can be combined by finding a common denominator. The result is ( \frac{ad + bc}{bd} ). Thus, ( \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} ).
If bd does not equal 0 then a over b plus c over d equals?
If bd ≠ 0, then a/b + c/d (the common denominator is bd) = (a x d)/(b x d) + (c x b)/(d x b) = ad/bd + cb/db = ad/bd + cb/bd = (ad + cb)/ bd
What is the difference between AD and BD?
AD means that Anno Domini , the year of lord and BD means that Before Christ
Is the difference of two rational numbers always a rational number?
Yes. Since they are rational numbers, let's call the first one a/b and the second one c/d where a,b,c, and d are integers. Now we can subtract by finding a common denominator. Let's use bd. So we have ad/bd-cb/bc= (ad-bc)/CD which is rational since we know ad and bc are integers being the product of integers and CD is also an integers. Call ad-bd=P and call CD=Q where P and Q are integers. We now see the difference is of two rationals is rational.
How many different ways can you write a fractions that has a numerator of 2 as a sum of fractions?
Infinitely many ways!Suppose you have the fraction 2/d.Pick any pair of integers a and b where b � 0.Then 2b-ad is and integer, as is bd so that (2b - ad)/bd is a fraction.Consider the fractions a/b and (2b - ad)/bdTheir sum isa/b + (2b-ad)/bd = ad/bd + (2b-ad)/bd = 2b/bd = 2/d - as required.Since a and b were chosen arbitrarily, there are infinitely many possible answers to the question.
Why the sum difference and product of 2 rational numbers rational?
A rational number is one that can be expressed as a/b The sum of two rational numbers is: a/b + c/d =ad/bd + bc/bd =(ad+bc)/bd =e/f which is rational The difference of two rational numbers is: a/b - c/d =ab/bd - bc/bd =(ab-bc)/bd =e/f which is rational The product of two rational numbers is: (a/b)(c/d) =ac/bd =e/f which is rational
Would the difference of a rational number and a rational number be rational?
The difference of two rational numbers is rational. Let the two rational numbers be a/b and c/d, where a, b, c, and d are integers. Any rational number can be represented this way. Their difference is a/b-c/d = ad/bd-cb/bd = (ad-cb)/bd. Products and differences of integers are always integers. This means that ad-cb is an integer, and so is bd. Thus, (ad-cb)/bd is a rational number (since it is the ratio of two integers). This is equivalent to the difference of the original two rational numbers.
What is the reason to ab plus bd equals ad?
The equation ( ab + bd = ad ) can be understood through factoring. The left side can be factored as ( b(a + d) ), which shows that it represents the product of ( b ) and the sum ( (a + d) ). However, this equation is not generally true unless specific conditions about the variables ( a ), ( b ), and ( d ) are met. In standard algebra, for ( ab + bd ) to equal ( ad ), we would require ( b = 0 ) or ( a + d = 0 ).
Have a rectangle bdfg with point a on side bd so that ba equals 16 cm and ad equals 8 cm what fraction of the area of the rectangle is inside triangle baf?
ba=(16/(16+8))bd=(16/24)bd=(2/3)bd area of the rectangle = bd*bf area of triangle = (2/3)bd*bf/2=(1/3)*bd*bf 1/3
What is the perimeter of the rectangle ABCD when AD equals 3 and BD equals 5?
It is 16 units.
How bad was black Sunday?
Really bad bd bd bd bd bd bd bd bd i dont know bd bd bd bd bd bd bd bd bd bd bd bd bd bd bd bd bd
Why is a over b plus c over d equal ad plus bc divided by bd always valid?
Because it can be derived from a large number of more basic principles applied to numbers. Since a/b and c/d are defined, then b and d are non-zero. Then a/b = ad/bd since multiplying the numerator and denominator of a fraction by a non-zero number leaves it unchanged. Similarly, c/d = bc/bd [These steps implicitly assume the commutative property of multiplication ie there is no difference between premultiplication and postmultiplication] Another property of numbers: If x = p and y = q then x+y = p+q [equals added to equals result in equals]. So a/b + c+d = ad/bd + bc/bd and finally, using the distributive property of multiplication [by 1/bd] over addition, = (ad + bc)/bd Maybe you wish you hadn't asked!
How do two fractions have a difference close to 0?
according to sum of fractions: a/b + c/d = (ad+bc)/bd hence for example: a=c=1 , b=d=2: (ad+bc)/bd = (2+2)/2x2 = 4/4 = 1 sagy
